Question

11. Suppose a force F varies as the product of two masses, m and M, inversely as
the square of the distance r between them. If F= 675 when m = 125, M = 225,
and r=0.0530, find the constant of proportionality k.
A. 61.48 X 10-10
B. 1.27X10-
C. 6.74 x 10-5
D. 6.76X10°
E. None

Answer 1

Answer:

$k = 6.74 * 10^{-5}$

Step-by-step explanation:

Given

$F\ \alpha\ \frac{m * M}{r^2}$ --- The variation

$F = 675; m = 125; M = 225; r = 0.0530$

Required

Determine the constant of proportionality (k)

$F\ \alpha\ \frac{m * M}{r^2}$

Express as an equation

$F = k\frac{m * M}{r^2}$

Multiply both sides by $r^2$

$Fr^2 = km * M$

Make k the subject

$k = \frac{Fr^2}{m * M}$

Given that: $F = 675; m = 125; M = 225; r = 0.0530$

We have:

$k = \frac{675* 0.0530^2}{125 * 225}$

$k = \frac{1.896075}{28125}$

$k = 0.000067416$

This can be represented as:

$k = 6.74 * 10^{-5}$